 Hi, I'm Zor. Welcome to Unizor education We continue solving problems combinatorics very good problems and I really like the combinatorics as a very good example of the mathematical problems which Build your logic build your analytical abilities. So the more the merrier All right. So Today I just have a few problems six of them All of them and the lecture itself are presented on the unizor.com website and that's where I suggest you to to view it from There are notes for this lecture. It's right on the side of the of the lecture itself of the video the problems are presented in the notes as well and The answer so you can first try to solve them yourself. That would be great if you do it and only then You could listen to this lecture So it's very important for you to spend some time thinking about these problems before you View the solution because again the purpose is development of your mind And the problem solving is the best way to do it Okay so Without further ado, let's just do the problems. Okay problem number one We have to prove that for any natural n and k number of combinations from n plus k by 2 and from a n plus k plus 1 by 2 represents a square of sum number All right. Well, this is a very simple exercise on Knowing the formulas now. I was talking about formulas in the previous lecture that you don't really have to remember them You can always derive the formulas of combinatorics just knowing the permutation n factorial formula. So anyway considering I know the formula and I'll just use it right now because to to prove it is really very simple. You can know the formula So this one is n plus k Factorial divided by 2 factorial and n plus k minus 2 Factorial now this one is n plus k plus 1 factorial divided by 2 factorial and n plus k plus 1 minus 2 So it would be minus 1 Factorial Well, obviously I have to bring them to the common denominator Now, let's just think about it. This one is one greater than this number, right? So if I will multiply both numerator and denominator by n plus k minus 1 n plus k minus 1 n plus k minus 1 What I will have is n plus k minus 1 and n plus k minus 2 factorial will give me n plus k minus 1 factorial So I have a common denominator this Now What do I have on the top? Well, let's just think about it n plus k factorial is all the numbers multiplied together from 1 to n plus k, right? So I think it would be beneficial for me if I will use n plus k minus 1 factorial here So these are all numbers from 1 to n plus k k minus 1 multiply together and n plus k and n plus k minus 1 Now, why do I need it? Obviously I want to cancel n plus n plus k minus 1 factorial Now, how about here? now again I would have it as n plus k minus 1 factorial and then next one would be n plus k and next one would be n plus k plus 1 So now I can factor out n plus k minus 1 factorial and in the parentheses, I will have n plus k times n plus k minus 1 plus n plus k times n plus k plus 1 divided by Well, 2 factorial is is true actually, right? So I don't need factorial minus 1 factorial and now we cancel n plus k minus 1 factorial now we factor out n plus k and what would be in the parentheses It would be n plus k minus 1 and n plus k plus 1. So it would be 2 n plus k divided by 2 So the result is n plus k square, which is exactly what needs to be proven So this sum is n plus k square and the problem next one Next one is we have to solve the equation partial permutations from x by x minus 3 equals x times permutation of x minus 2 Well again, we have to use the formula which again I happen to remember Now this one is factorial of the x divided by factorial of the difference x minus x minus 3. So it's 3 factorial Now this one is x times x minus 2 factorial, right? Now 3 factorial is 1 times 2 times 3 is 6 So we have x square Sorry x factorial equals to 6x x minus 2 factorial Now what is x factorial is the product of all numbers from 1 to x And this is from 1 to x minus 2. So it's 2 less than to x, right? So I can basically Say that this is x minus 2 factorial times x minus 1 And times x that's what x factorial is all the numbers from 1 to x minus 2 Then x minus 1 and then x and this is equal to 6x x minus 2 factorial And now we can cancel this out We are not talking about anything equal to 0 in this particular case because it's factorials, right? Zero factorial for instance is 1 anyway. So we obviously are talking about numbers greater than 2 actually greater than 3 um, maybe equal That's the condition of this particular Equation we have to really think about this this way Um, otherwise we will have negative number Which which doesn't make much much sense anymore. Now x is also canceling out So what do I have x minus 1 equals 6 x equals 5? Uh, sorry 7 7 That's the solution So for those people who really Need to make sure this is a solution Substitute it here And what do you have p from 7 to 4 Should be equal to 7 times p 5 right So this is 7 factorial divided by 7 minus 4 3 factorial Should be equals to 7 times 5 factorial Well, let's just count 7 factorial is 1 2 3 4 5 6 7 Which is 5 factorial is 120 6 factorial is 720 And 7 factorial is you have to multiply it by 7 4 50 I think that's where it is And we have to divide it by 3 factorial which is 6 6 which is 8 2 4 2 4 4 8 40. So this is 8 40 How about this 5 factorial is 120 times 7 8 40. Okay, so Checking is fine So the root of this equation is 7. That's the solution Next and obviously it's greater than 3 Okay, now we have to prove That number of permutations is equal to n minus 1 The number of permutations of n minus 1 and n minus 2 Well, this is probably one of the simplest problems because on the left I have n factorial right on the right. I have n minus 1 factorial, which is this one Plus n minus 2 factorial Which is this one and I multiply then by n minus 1 Which is equal to n minus 1 Now what is n minus 1 factorial? You can write it as n minus 1 and n minus 2 factorial, right? Because this is all the numbers from 1 to n minus 2 And n minus 1 gives you n minus 1 factorial plus n minus 2 factorial Now we can factor out n minus 2 factorial And I will have this And what do I have here? n minus 1 Plus 1 So what do I have? All together n minus 2 factorial, which is this one Then n minus 1, which is this one And then n So which is numbers from 1 to n minus 2 and n minus 1 and n which is n factorial Which is something which we're supposed to get Simple Next Okay We have certain Building co-op building and it has board of directors So this cooperative has seven members Of the board Now there are two positions which they have to basically Choose candidates among themselves for one position is president of the board And the second position is I mean it's also one And one position treasurer Now question is now the rest or other five people members of the board are just members of the board Now my question is how many different distributions of responsibilities this board can have Considering, you know, we have Seven people and these two positions must be chosen. Well, there are many ways to do this type of This type of problem. For instance We have seven different choices to choose a president, right? We have seven people So we have seven different choices to choose a president. So it's seven With with each of those choices We have remaining six people and we have to choose one of those for a treasurer And there are six choices, obviously. So that makes it for you to That's some kind of a pure logical explanation without any formulas, etc On the other hand, you can think about more like Formal combinatorics What does it mean to choose the president and the treasurer out of the seven people? Well, it means you have to really Make choice of partial permutations of Seven by two So you have to choose two, but the order is important like the first one is a president The second one is a treasurer. So you have to choose two people in certain order That's what partial permutation actually is all about and the formula is Seven factorial divided by seven minus two factorial Which is seven factorial divided by five factorial, which is seven times six Which is four two two And probably there are some other ways to approach this problem But these two are sufficient at least So we have four two two Next is A System of two equations combinatorial equations with two variables unknown Well, we have not solved Linear systems of linear equation for a very long time since I was doing algebra lessons. So anyway Well, the easiest way to approach this is the following First of all, let's just do this type of New variables now In this new language with new variables. I have something which is very easy to solve Now how can I solve this well subtract from the first one I subtract the second one I will have five v equals 50 So v is 10 And u is If this is 10, this is 30 60 six 10 I meant So 60 is that right? 60 minus 20 40 60 plus 30 90, right? Okay, so v is equal 10, u is equal 60 u is equal 16 And v is equal to 10 That's what we've got. All right. That's good Now Let's just do it this way x factorial divided by x minus y factorial equals 60 x factorial divided by x factorial and x minus y factorial equals 10 Okay, now let's compare one to another. What do we have? If I will divide this by this My x factorial will cancel out and x minus y factorial will cancel out This is one over x factorial, but we divide on this, right? So it will be uh in the numerator. So I will have only x factorial and 60 divided by 10 gives me 6 Well, that defines x x is equal to 3 right 3 factorial is 6 Well, once I have x now I can actually use something like this um Oh wait a moment. I think I made a mistake. This should be y factorial. Sorry y factorial So it's y is equal to 6 3 x factorial divided by y factorial and x minus. Yes, that's right It's my mistake. Anyway, so if I divide this by this I will have y factorial is equal to 6 and y is equal to 3 correct Now since I know that y is equal to 3 Let's just do this job for for x We can take any one of those two equations for instance this one and substitute 3 So x factorial divided by x minus 3 factorial equals 60 Now what is x factorial relative to x minus minus 3 factorial? Well, I can always Write x factorial as x minus 3 factorial And x minus 2 x minus 1 and x right That's what will be in my numerator and now Obviously I can do this So what do I have? I have x times x minus 1 times x minus 2 equals 60 Now we definitely know these are integer numbers positive integer numbers So we have three numbers in a row multiplied Give give you 60 Well, you don't want to to to solve the the cubic equation obviously The best way is just to find a couple of numbers and think about These numbers are not supposed to be very large anyway. So let's say it's 5 4 and 3. What happens 5 times 4 20 times 360. So I just guessed my my first guess was x is equal to 5 So that's basically your Best approach in this particular case Considering the numbers are relatively small and you definitely know that these numbers are integers. So that's the solution Okay, and my last Problem today is Okay, you have two male men and you have n two people Male men and you have n letters Which are supposed to be delivered to 10 different addresses Now they have to split the job somehow among themselves Question is how many different ways of this distribution of the job exists? Well, there are two different approaches to do this approach number one Basically, we have to pick something pick the job for one of them and that actually defines the job for another Right. So what kind of a job I can pick for the first one? Well, I can either choose For him to deliver zero letters out of the n or one letter out of the n Or two etc or all n letters to be delivered by this particular person Now how many different combinations to choose zero letters from n? Well, it's c from n to zero This is c from n to one. This is from c from n to two So it's number of combinations of two letters out of n So I have to pick two letters and say, okay, this is your job Or I have to pick one letter and say this is your job. So the sum of these Give me the number of different job distributions among these two guys. So this is the answer great It's a very long formula, right? Well, but let's do it differently Let's pick one letter out of this n And assign it to one or the other mailman. There are two choices, right? So we have two choices Let's pick up another letter The second one Well, it's also two choices And with each one of the previous choices, we can have each one of these choices I have to multiply the choices, right? And then the third letter, the fourth letter, etc All letters are supposed to be distributed among these two Using the same logic and obviously since choices are multiplying, I have to multiply them. So it's two to the n's degree Look at this formula This is quite an interesting formula Well, if you don't recognize it I can tell you this is a newtons binomial So if you use newtons binomial Which I actually am presenting on unizer.com among the lectures on induction I think it's the problem 1.6 in the induction. That's from from the beginning of this course where Some math concepts actually are explained In particular the concept of mathematical induction. So that's where actually I'm presenting the binomial newtons binomial and and suggest the Proof of the formula by induction. So if you will open up this particular formula binomial Newtons binomial, you will get exactly this And this is two to the n's degree So that's an interesting Confirmation if you wish of the newtons binomial formula in case one plus one So the two to the n's degree is the sum of all these binomial coefficients as they are called Okay, that was my last problem I do suggest you to go through these problems again on the unizer.com website Try to do it yourself and then read the solutions which I am providing and there are answers So you can check your solutions Well, basically, that's it. Thank you very much and good luck